Question

Two parallel plate capacitors each of 15 µF capacity are connected in series. The space between the plates of one capacitor is filled with a dielectric material of dielectric constant \( K = 2 \). The equivalent capacitance of the system will be:

• A. 45 µF
• B. 30 µF
• C. 10 µF
• D. 15 µF

Hint:

When capacitors are connected in series, the equivalent capacitance is always less than the smallest capacitance. If a dielectric is inserted, the capacitance increases for that specific capacitor.

Answer 1

The Correct Option is C

Step 1: Understand the given information.
We are given two capacitors, each with a capacitance of \( 15 \, \mu{F} \). The space between the plates of one of the capacitors is filled with a dielectric material with a dielectric constant \( K = 2 \). These two capacitors are connected in series.
Step 2: Capacitance of a capacitor with dielectric.
When a dielectric material is inserted between the plates of a capacitor, the capacitance \( C' \) is increased by a factor of \( K \), the dielectric constant. The capacitance of the capacitor with the dielectric material is:
\[ C_{{dielectric}} = K \times C = 2 \times 15 \, \mu{F} = 30 \, \mu{F} \]
Step 3: Equivalent capacitance for capacitors in series.
The equivalent capacitance \( C_{{eq}} \) for two capacitors \( C_1 \) and \( C_2 \) connected in series is given by:
\[ \frac{1}{C_{{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Where:
\( C_1 = 15 \, \mu{F} \) (capacitor without the dielectric),
\( C_2 = 30 \, \mu{F} \) (capacitor with the dielectric).
Substitute the values into the formula:
\[ \frac{1}{C_{{eq}}} = \frac{1}{15} + \frac{1}{30} \]
\[ \frac{1}{C_{{eq}}} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} \]
Thus:
\[ C_{{eq}} = \frac{30}{3} = 10 \, \mu{F} \]
Step 4: Conclusion.
The equivalent capacitance of the system is 10 µF, which corresponds to option (2).